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Consider an element $X\in\mathfrak g$ of some Lie algebra $\mathfrak g$. I understand that $\mathfrak g$ can be represented via its action on other elements of the same algebra, as $\operatorname{ad}(X)Y\equiv[X,Y]$, so that $\operatorname{ad}(X)\in\operatorname{GL}(\mathfrak g)$.

One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $\operatorname{ad}(X)$.

However, I usually don't see any direct mention of the inner product with respect to which these things are defined. To properly define what something such as $\operatorname{ad}(X)_{ij}$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $\{X_i\}$ the elements of some basis for the algebra)?

Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?

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    $\begingroup$ Trace of a matrix is independent of the basis. $\endgroup$ – Moishe Kohan Mar 29 '19 at 17:15
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    $\begingroup$ One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. math.stackexchange.com/q/72303/96384. $\endgroup$ – Torsten Schoeneberg Mar 29 '19 at 17:16
  • $\begingroup$ @MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace $\endgroup$ – glS Mar 29 '19 at 17:17
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    $\begingroup$ Surely $\mathfrak{g}$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it? $\endgroup$ – Torsten Schoeneberg Mar 29 '19 at 17:27
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    $\begingroup$ I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^{-1})=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see math.stackexchange.com/questions/1369839/… $\endgroup$ – Moishe Kohan Mar 29 '19 at 18:18
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Consider $\mathfrak{sl}_2$ with basis $\{e,f,h\}$. You have $$ad(e)(e)=0,\;\;\;ad(e)(f)=h,\;\;\;ad(e)(h)=-2e$$ so the matrix of $ad(e)$ in this basis is $$ \begin{pmatrix}0&0&-2\\0&0&0\\0&1&0\end{pmatrix} $$ which has trace 0.

You can check that the matrix for $ad(h)$ is $$ \begin{pmatrix}2&0&0\\0&-2&0\\0&0&0\end{pmatrix} $$ which again has trace 0. Can you do $ad(f)$?

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  • $\begingroup$ You also need to know that the trace is independent of a choice of basis. $\endgroup$ – Qiaochu Yuan Mar 29 '19 at 17:35
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    $\begingroup$ That has been discussed at length in the comments and I take that as an elementary linear algebra fact. @QiaochuYuan $\endgroup$ – David Hill Mar 29 '19 at 17:57

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